【leetcode】923. 3Sum With Multiplicity

题目如下：

Given an integer array A, and an integer target, return the number of tuples i, j, k  such that i < j < k and A[i] + A[j] + A[k] == target.

As the answer can be very large, return it modulo 10^9 + 7.

Example 1:

Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (A[i], A[j], A[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Example 2:

Input: A = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
A[i] = 1, A[j] = A[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Note:

1. 3 <= A.length <= 3000
2. 0 <= A[i] <= 100
3. 0 <= target <= 300

解题思路：虽然A.length 最大值是300，但是A[i]的值在0~100之间，说明A中有很多重复值，对A去重后length最大值也就100，所以O(n^3)的复杂度完全可以接受。首先对A去重，假设A[i] * A[j] * A[k] == target (i<=j<=k)，那么 A[i] 、A[j]、A[k] 三者之间的值有这么几种情况:

a.三者相等： 这种情况，一共存在C(A[i]在A中的个数,3)种组合 (A[i]在A中的个数 >= 3, 这个表达的是A[i]在去重前的A出现的次数)

b.任意两者相等：假设A[i] == A[j] != A[k] ,那么一共存在 C(A[i]在A中的个数,2) * A[k]在A中出现的次数 （A[i]在A中的个数,2) >= 2）

c.三者完全不同：这个最简单,一共存在 A[i]在A中出现的次数 * A[j]在A中出现的次数 * A[k]在A中出现的次数

代码如下：

class Solution(object):
    def threeSumMulti(self, A, target):
        """
        :type A: List[int]
        :type target: int
        :rtype: int
        """
        def combination(n,m):
            v1 = 1
            times = 0
            while times < m:
                v1 *= n
                n -= 1
                times += 1
            v2 = 1
            while m > 0:
                v2 *= m
                m -= 1
            return v1 / v2

        dic = {}
        for i in A:
            dic[i] = dic.setdefault(i, 0) + 1
        ul = list(set(A))
        res = 0
        for i in range(len(ul)):
            for j in range(i,len(ul)):
                for k in range(j,len(ul)):
                    if (ul[i] + ul[j] + ul[k]) != target:
                        continue
                    elif ul[i] == ul[j] == ul[k]:
                        if dic[ul[i]] >= 3:
                            res += combination(dic[ul[i]],3)
                    elif ul[i] == ul[j]:
                        if dic[ul[i]] >= 2:
                            res += (combination(dic[ul[i]],2) * dic[ul[k]])
                    elif ul[i] == ul[k]:
                        if dic[ul[i]] >= 2:
                            res += (combination(dic[ul[i]], 2) * dic[ul[j]])
                    elif ul[j] == ul[k]:
                        if dic[ul[j]] >= 2:
                            res += (combination(dic[ul[j]], 2) * dic[ul[i]])
                    else:
                        res +=  (dic[ul[i]] * dic[ul[j]] * dic[ul[k]])
        return res % (pow(10,9) + 7)